Problem: $\lim_{x\to -1}\dfrac{x^4+2x^3+x^2}{x+1}=$
Solution: Substituting $x=-1$ into $\dfrac{x^4+2x^3+x^2}{x+1}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression on our hands, let's try to simplify it. $\dfrac{x^4+2x^3+x^2}{x+1}$ can be simplified as $x^3+x^2$, for $x\neq -1$. This means that the two expressions have the same value for all $x$ -values (in their domains) except for $-1$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{x^4+2x^3+x^2}{x+1}=x^3+x^2$ for all $x$ -values in the interval $(-2,0)$ except for $x=-1$. Therefore, $\lim_{x\to -1}\dfrac{x^4+2x^3+x^2}{x+1}=\lim_{x\to -1}(x^3+x^2)=0$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to -1}\dfrac{x^4+2x^3+x^2}{x+1}=0$.